See this tensorflow-probability issue
tensorflow==2.7.0
tensorflow-probability==0.14.1
TLDR
To perform VI on discrete RVs, should I use:
A- the REINFORCE gradient estimator
B- the Gumbel-Softmax reparametrization
C- another solution
and how to implement it ?
Problem statement
Sorry in advance for the long issue, but I believe the problem requires some explaining.
I want to implement a Hierarchical Bayesian Model involving both continuous and discrete Random Variables. A minimal example is a Gaussian Mixture model:
import tensorflow as tf
import tensorflow_probability as tfp
tfd = tfp.distributions
tfb = tfp.bijectors
G = 2
p = tfd.JointDistributionNamed(
model=dict(
mu=tfd.Sample(
tfd.Normal(0., 1.),
sample_shape=(G,)
),
z=tfd.Categorical(
probs=tf.ones((G,)) / G
),
x=lambda mu, z: tfd.Normal(
loc=mu[z],
scale=1.
)
)
)
In this example I don't use the tfd.Mixture API on purpose to expose the Categorical label. I want to perform Variational Inference in this context, and for instance given an observed x fit over the posterior of z a Categorical distribution with parametric probabilities:
q_probs = tfp.util.TransformedVariable(
tf.ones((G,)) / G,
tfb.SoftmaxCentered(),
name="q_probs"
)
q_loc = tf.Variable(0., name="q_loc")
q_scale = tfp.util.TransformedVariable(
1.,
tfb.Exp(),
name="q_scale"
)
q = tfd.JointDistributionNamed(
model=dict(
mu=tfd.Normal(q_loc, q_scale),
z=tfd.Categorical(probs=q_probs)
)
)
The issue is: when computing the ELBO and trying to optimize for the optimal q_probs I cannot use the reparameterization gradient estimators: this is AFAIK because z is a discrete RV:
def log_prob_fn(**kwargs):
return p.log_prob(
**kwargs,
x=tf.constant([2.])
)
optimizer = tf.optimizers.SGD()
#tf.function
def fit_vi():
return tfp.vi.fit_surrogate_posterior(
target_log_prob_fn=log_prob_fn,
surrogate_posterior=q,
optimizer=optimizer,
num_steps=10,
sample_size=8
)
_ = fit_vi()
# This last line raises:
# ValueError: Distribution `surrogate_posterior` must be reparameterized, i.e.,a diffeomorphic transformation
# of a parameterless distribution. (Otherwise this function has a biased gradient.)
I'm looking into a way to make this work. I've identified at least 2 ways to circumvent the issue: using REINFORCE gradient estimator or the Gumbel-Softmax reparameterization.
A- REINFORCE gradient
cf this TFP API link a classical result in VI is that the REINFORCE gradient can deal with a non-differentiable objective function, for instance due to discrete RVs.
I can use a tfp.vi.GradientEstimators.SCORE_FUNCTION estimator instead of the tfp.vi.GradientEstimators.REPARAMETERIZATION one using the lower-level tfp.vi.monte_carlo_variational_loss function?
Using the REINFORCE gradient, In only need the log_prob method of q to be differentiable, but the sample method needn't be differentiated.
As far as I understood it, the sample method for a Categorical distribution implies a gradient break, but the log_prob method does not. Am I correct to assume that this could help with my issue? Am I missing something here?
Also I wonder: why is this possibility not exposed in the tfp.vi.fit_surrogate_posterior API ? Is the performance bad, meaning is the variance of the estimator too large for practical purposes ?
B- Gumbel-Softmax reparameterization
cf this TFP API link I could also reparameterize z as a variable y = tfd.RelaxedOneHotCategorical(...) . The issue is: I need to have a proper categorical label to use for the definition of x, so AFAIK I need to do the following:
p_GS = tfd.JointDistributionNamed(
model=dict(
mu=tfd.Sample(
tfd.Normal(0., 1.),
sample_shape=(G,)
),
y=tfd.RelaxedOneHotCategorical(
temperature=1.,
probs=tf.ones((G,)) / G
),
x=lambda mu, y: tfd.Normal(
loc=mu[tf.argmax(y)],
scale=1.
)
)
)
...but his would just move the gradient breaking problem to tf.argmax. This is where I maybe miss something. Following the Gumbel-Softmax (Jang et al., 2016) paper, I could then use the "STRAIGHT-THROUGH" (ST) strategy and "plug" the gradients of the variable tf.one_hot(tf.argmax(y)) -the "discrete y"- onto y -the "continuous y".
But again I wonder: how to do this properly ? I don't want to mix and match the gradients by hand, and I guess an autodiff backend is precisely meant to avoid me this issue. How could I create a distribution that differentiates the forward direction (sampling a "discrete y") from the backward direction (gradient computed using the "continuous y") ? I guess this is the meant usage of the tfd.RelaxedOneHotCategorical distribution, but I don't see this implemented anywhere in the API.
Should I implement this myself ? How ? Could I use something in the lines of tf.custom_gradient?
Actual question
Which solution -A or B or another- is meant to be used in the TFP API, if any? How should I implement said solution efficiently?
So the ides was not to make a Q&A but I looked into this issue for a couple days and here are my conclusions:
solution A -REINFORCE- is a possibility, it doesn't introduce any bias, but as far as I understood it it has high variance in its vanilla form -making it prohibitively slow for most real-world tasks. As detailed a bit below, control variates can help tackle the variance issue;
solution B, Gumbell-Softmax, exists as well in the API, but I did not find any native way to make it work for hierarchical tasks. Below is my implementation.
First off, we need to reparameterize the joint distribution p as the KL between a discrete and a continuous distribution is ill-defined (as explained in the Maddison et al. (2017) paper). To not break the gradients, I implemented a simple one_hot_straight_through operation that converts the continuous RV y into a discrete RV z:
G = 2
#tf.custom_gradient
def one_hot_straight_through(y):
depth = y.shape[-1]
z = tf.one_hot(
tf.argmax(
y,
axis=-1
),
depth=depth
)
def grad(upstream):
return upstream
return z, grad
p = tfd.JointDistributionNamed(
model=dict(
mu=tfd.Sample(
tfd.Normal(0., 1.),
sample_shape=(G,)
),
y=tfd.RelaxedOneHotCategorical(
temperature=1.,
probs=tf.ones((G,)) / G
),
x=lambda mu, y: tfd.Normal(
loc=tf.reduce_sum(
one_hot_straight_through(y)
* mu
),
scale=1.
)
)
)
The variational distribution q follows the same reparameterization and the following code bit does work:
q_probs = tfp.util.TransformedVariable(
tf.ones((G,)) / G,
tfb.SoftmaxCentered(),
name="q_probs"
)
q_loc = tf.Variable(tf.zeros((2,)), name="q_loc")
q_scale = tfp.util.TransformedVariable(
1.,
tfb.Exp(),
name="q_scale"
)
q = tfd.JointDistributionNamed(
model=dict(
mu=tfd.Independent(
tfd.Normal(q_loc, q_scale),
reinterpreted_batch_ndims=1
),
y=tfd.RelaxedOneHotCategorical(
temperature=1.,
probs=q_probs
)
)
)
def log_prob_fn(**kwargs):
return p.log_prob(
**kwargs,
x=tf.constant([2.])
)
optimizer = tf.optimizers.SGD()
#tf.function
def fit_vi():
return tfp.vi.fit_surrogate_posterior(
target_log_prob_fn=log_prob_fn,
surrogate_posterior=q,
optimizer=optimizer,
num_steps=10,
sample_size=8
)
_ = fit_vi()
Now there are several issues with that design:
first off we needed to reparameterize not only q but also p so we "modify our target model". This results in our models p and q not outputing discrete RVs like originally intended but continuous RVs. I think that the introduction of a hard option like in the torch implem could be a nice addition to overcome this issue;
second we introduce the burden of setting up the temperature parameter. The latter make the continuous RV y smoothly converge to its discrete counterpart z. An annealing strategy, reducing the temperature to reduce the bias introduced by the relaxation at the cost of a higher variance can be implemented. Or the temperature can be learned online, akin to an entropy regularization (see Maddison et al. (2017) and Jang et al. (2017));
the gradient obtained with this estimator are biased, which probably can be acceptable for most applications but is an issue in theory.
Recent methods like REBAR (Tucker et al. (2017)) or RELAX (Grathwohl et al. (2018)) can instead obtain unbiased estimators with a lower variance than the original REINFORCE. But they do so at the cost of introducing -learnable- control variates with separate losses. Modifications of the one_hot_straight_through functions could probably implement this.
In conclusion my opinion is that the tensorflow probability support for discrete RVs optimization is too scarce at the moment and that the API lacks native functions and tutorials to make it easier for the user.
Related
I'm trying to understand the variational inference module in tensorflow; I have a particular use case I'm hoping to use it for.
I want to make a custom distribution, the RV of which is a transformation of a vector of independent gamma RV's. This transformation removes one degree of freedom.
For simplicity's sake, let's consider the Dirichlet distribution. If x is an independent gamma vector with shape parameter vector a, then y = x / sum(x) is a dirichlet vector with the same shape vector, and sums to 1. Thus it loses 1 degree of freedom in the transformation.
Let's say I want to implement this distribution as a tfp.distributions.TransformedDistribution. Would that be possible? The Bijector class assumes implementation of both forward and inverse transformations, which, after the sum is integrated out, is no longer possible.
How would I go about implementing the Dirichlet in TransformedDistribution?
If my latent representation of variational autoencoder(VAE) is r, and my dataset is x, does vae's latent representation follows normalization based on r or x?
If r= 10, that means it has 10 means and variance (multi-gussain) and distribution comes from data whole data x?
Or r = 10 constructs one distribution based on r, and every sample try to follow this distribution
I'm confused about which one is correct
VAE constructs a mapping e(x) -> Z (encoder), and d(z) -> X (decoder). This means that every elements of your input space x will be mapped through an encoder e(x) into a single, r-dimensional Gaussian. It is not a "mixture", it is just a single gaussian with diagonal covariance matrix.
I'll add my 2 cents to #lejlot answer.
Your encoder in VAE will map your sample to a distribution, that in your case has 10 dimensions... that distribution is used to say "ok my best estimate of this property of this sample is mu, but I'm not too sure, so consider that it might have variance sigma"
Therefore, you have a distribution for each sample.
However, in order to make sampling easier in VAE, we ask the VAE to keep the distributions as close to a known one, that is the standard normal distribution (we know "where the distributions are located", if you check the latent space in a normal AE you will see that you will have groups far from eachother).
I'm new to automatic differentiation programming, so this maybe a naive question. Below is a simplified version of what I'm trying to solve.
I have two input arrays - a vector A of size N and a matrix B of shape (N, M), as well a parameter vector theta of size M. I define a new array C(theta) = B * theta to get a new vector of size N. I then obtain the indices of elements that fall in the upper and lower quartile of C, and use them to create a new array A_low(theta) = A[lower quartile indices of C] and A_high(theta) = A[upper quartile indices of C]. Clearly these two do depend on theta, but is it possible to differentiate A_low and A_high w.r.t theta?
My attempts so far seem to suggest no - I have using the python libraries of autograd, JAX and tensorflow, but they all return a gradient of zero. (The approaches I have tried so far involve using argsort or extracting the relevant sub-arrays using tf.top_k.)
What I'm seeking help with is either a proof that the derivative is not defined (or cannot be analytically computed) or if it does exist, a suggestion on how to estimate it. My eventual goal is to minimize some function f(A_low, A_high) wrt theta.
This is the JAX computation that I wrote based on your description:
import numpy as np
import jax.numpy as jnp
import jax
N = 10
M = 20
rng = np.random.default_rng(0)
A = jnp.array(rng.random((N,)))
B = jnp.array(rng.random((N, M)))
theta = jnp.array(rng.random(M))
def f(A, B, theta, k=3):
C = B # theta
_, i_upper = lax.top_k(C, k)
_, i_lower = lax.top_k(-C, k)
return A[i_lower], A[i_upper]
x, y = f(A, B, theta)
dx_dtheta, dy_dtheta = jax.jacobian(f, argnums=2)(A, B, theta)
The derivatives are all zero, and I believe this is correct, because the change in value of the outputs does not depend on the change in value of theta.
But, you might ask, how can this be? After all, theta enters into the computation, and if you put in a different value for theta, you get different outputs. How could the gradient be zero?
What you must keep in mind, though, is that differentiation doesn't measure whether an input affects an output. It measures the change in output given an infinitesimal change in input.
Let's use a slightly simpler function as an example:
import jax
import jax.numpy as jnp
A = jnp.array([1.0, 2.0, 3.0])
theta = jnp.array([5.0, 1.0, 3.0])
def f(A, theta):
return A[jnp.argmax(theta)]
x = f(A, theta)
dx_dtheta = jax.grad(f, argnums=1)(A, theta)
Here the result of differentiating f with respect to theta is all zero, for the same reasons as above. Why? If you make an infinitesimal change to theta, it will in general not affect the sort order of theta. Thus, the entries you choose from A do not change given an infinitesimal change in theta, and thus the derivative with respect to theta is zero.
Now, you might argue that there are circumstances where this is not the case: for example, if two values in theta are very close together, then certainly perturbing one even infinitesimally could change their respective rank. This is true, but the gradient resulting from this procedure is undefined (the change in output is not smooth with respect to the change in input). The good news is this discontinuity is one-sided: if you perturb in the other direction, there is no change in rank and the gradient is well-defined. In order to avoid undefined gradients, most autodiff systems will implicitly use this safer definition of a derivative for rank-based computations.
The result is that the value of the output does not change when you infinitesimally perturb the input, which is another way of saying the gradient is zero. And this is not a failure of autodiff – it is the correct gradient given the definition of differentiation that autodiff is built on. Moreover, were you to try changing to a different definition of the derivative at these discontinuities, the best you could hope for would be undefined outputs, so the definition that results in zeros is arguably more useful and correct.
Let z is a complex variable, C(z) is its conjugation.
In complex analysis theory, the derivative of C(z) w.r.t z don't exist. But in tesnsorflow, we can calculate dC(z)/dz and the result is just 1.
Here is an example:
x = tf.placeholder('complex64',(2,2))
y = tf.reduce_sum(tf.conj(x))
z = tf.gradients(y,x)
sess = tf.Session()
X = np.random.rand(2,2)+1.j*np.random.rand(2,2)
X = X.astype('complex64')
Z = sess.run(z,{x:X})[0]
The input X is
[[0.17014372+0.71475762j 0.57455420+0.00144318j]
[0.57871044+0.61303568j 0.48074263+0.7623235j ]]
and the result Z is
[[1.-0.j 1.-0.j]
[1.-0.j 1.-0.j]]
I don't understand why the gradient is set to be 1?
And I want to know how tensorflow handles the complex gradients in general.
How?
The equation used by Tensorflow for the gradient is:
Where the '*' means conjugate.
When using the definition of the partial derivatives wrt z and z* it uses Wirtinger Calculus. Wirtinger calculus enables to calculate the derivative wrt a complex variable for non-holomorphic functions. The Wirtinger definition is:
Why this definition?
When using for example Complex-Valued Neural Networks (CVNN) the gradients will be used over non-holomorphic, real-valued scalar function of one or several complex variables, tensorflow definition of a gradient can then be written as:
This definition corresponds with the literature of CVNN like for example chapter 4 section 4.3 of this book or Amin et al. (between countless examples).
Bit late, but I came across this issue recently too.
The key point is that TensorFlow defines the "gradient" of a complex-valued function f(z) of a complex variable as "the gradient of the real map F: (x,y) -> Re(f(x+iy)), expressed as a complex number" (the gradient of that real map is a vector in R^2, so we can express it as a complex number in the obvious way).
Presumably the reason for that definition is that in TF one is usually concerned with gradients for the purpose of running gradient descent on a loss function, and in particular for identifying the direction of maximum increase/decrease of that loss function. Using the above definition of gradient means that a complex-valued function of complex variables can be used as a loss function in a standard gradient descent algorithm, and the result will be that the real part of the function gets minimised (which seems to me a somewhat reasonable interpretation of "optimise this complex-valued function").
Now, to your question, an equivalent way to write that definition of gradient is
gradient(f) := dF/dx + idF/dy = conj(df/dz + dconj(f)/dz)
(you can easily verify that using the definition of d/dz). That's how TensorFlow handles complex gradients. As for the case of f(z):=conj(z), we have df/dz=0 (as you mention) and dconj(f)/dz=1, giving gradient(f)=1.
I wrote up a longer explanation here, if you're interested: https://github.com/tensorflow/tensorflow/issues/3348#issuecomment-512101921
There are few key parameters associated with Linear Regression e.g. Adjusted R Square, Coefficients, P-value, R square, Multiple R etc. While using google Tensorflow API to implement Linear Regression how are these parameter mapped? Is there any way we can get the value of these parameters after/during model execution
From my experience, if you want to have these values while your model runs then you have to hand code them using tensorflow functions. If you want them after the model has run you can use scipy or other implementations. Below are some examples of how you might go about coding R^2, MAPE, RMSE...
total_error = tf.reduce_sum(tf.square(tf.sub(y, tf.reduce_mean(y))))
unexplained_error = tf.reduce_sum(tf.square(tf.sub(y, prediction)))
R_squared = tf.sub(tf.div(total_error, unexplained_error),1.0)
R = tf.mul(tf.sign(R_squared),tf.sqrt(tf.abs(unexplained_error)))
MAPE = tf.reduce_mean(tf.abs(tf.div(tf.sub(y, prediction), y)))
RMSE = tf.sqrt(tf.reduce_mean(tf.square(tf.sub(y, prediction))))
I believe the formula for R2 should be the following. Note that it would go negative when the network is so bad that it does a worse job than the mere average as a predictor:
total_error = tf.reduce_sum(tf.square(tf.subtract(y, tf.reduce_mean(y))))
unexplained_error = tf.reduce_sum(tf.square(tf.subtract(y, pred)))
R_squared = tf.subtract(1.0, tf.divide(unexplained_error, total_error))
Adjusted_R_squared = 1 - [ (1-R_squared)*(n-1)/(n-k-1) ]
whereas n is the number of observations and k is the number of features.
You should not use a formula for R Squared. This exists in Tensorflow Addons. You will only need to extend it to Adjusted R Squared.
I would strongly recommend against using a recipe to calculate r-squared itself! The examples I've found do not produce consistent results, especially with just one target variable. This gave me enormous headaches!
The correct thing to do is to use tensorflow_addons.metrics.RQsquare(). Tensorflow Add Ons is on PyPi here and the documentation is a part of Tensorflow here. All you have to do is set y_shape to the shape of your output, often it is (1,) for a single output variable.
Then you can use what RSquare() returns in your own metric that handled the adjustments.